🚀 go-pugleaf

In article <1992Nov2.205433.21208@tamsun.tamu.edu> cmenzel@tamsun.tamu.edu (Chris Menzel) writes:
>In article <1992Nov2.040843.4524@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
>>That Russell felt obliged to give a more broadly applicable argument
>>suggests that FA was not held to be an evident property of sets by
>>mathematicians in 1905.
>
>Most certainly true.  Well-foundedness wasn't really an issue until
>the cumulative hierarchy started to take hold.  Mirimanoff explored
>the axiom a bit around 1918, and sensed an important distinction
>between well-founded and non-well-founded ("extraodinary," in his
>terms) sets.  But it wasn't really until Zermelo's great 1931 paper
>(still untranslated, for some reason) "Uber Grenzzahlen und
>Mengenbereiche" in which he described the hierarchy in great detail
>that foundation became an unquestioned part of set theory.
>
>>That it is self-evident to you now seems more
>>an accident of culture---your training---than of something inherent in
>>the notion of set....
>
>I think that's too strong if, like me, you were introduced to set
>theory via the iterative hierarchy.  If that's your picture of sets,
>then the axiom *is* self-evident, and well-foundedness *is* an
>inherent property of sets.  The question is rather whether the
>iterative hierarchy encompasses *all* the sets, or whether we've just
>been looking at a rather limited portion of the universe, as is the
>case if we take the universe to be nonwellfounded ala Aczel.
>

Actually, Aczel's universe is not very much larger than the usual
universe.  It is hierarchical in much the same way as the universe of
ZFC, with the stages H(a) of the hierarchy generated as follows:

	H(0) = {{}}

	H(a + 1) = sets with the property that membership on their
transitive closure is isomorphic to a graph with nodes taken from
H(a).

	H(L) = U{a < L}[H(a)] (L limit).

	Under AFA, the isomorphism type of the graph of membership on
the transitive closure of a set determines the set uniquely, so the
cardinality of H(a + 1) can be shown to be the same as the cardinality
of the power set of H(a) for infinite a (this is not true for finite
a), which implies that the cardinalities of the H(a)'s are the same as
those of the V(a)'s in the well-founded universe, for infinite
ordinals a.

	AFA does not actually make the universe noticeably larger.


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu